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  • A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight , long and at right angle. At D wire forms  circular turn DMND of radius R . AB, BC parts are tangential to circular turn at N and D . Magnetic field at t

A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight , long and at right angle. At D wire forms  circular turn DMND of radius R . AB, BC parts are tangential to circular turn at N and D . Magnetic field at the centre of circle is :
Option: 1   \frac{\mu _{0}I}{2\pi R}\left ( \pi + \frac{1}{\sqrt{2}}\right )

Option: 2  \frac{\mu _{0}I}{2 R}

Option: 3 \frac{\mu _{0}I}{2\pi R}\left ( \pi + 1 \right )

Option: 4 \frac{\mu _{0}I}{2\pi R}\left ( \pi - \frac{1}{\sqrt{2}}\right )
 

Answers (1)

best_answer

 

 

As

Magnetic Field due to current in straight wire -

  B=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{r}\left(\sin \phi_{1}+\sin \phi_{2}\right)   

  

So 

 

Magnitude of net magnetic field at O is

\\B_O=B_1+B_2+B_3\\=-\frac{\mu_0i}{4\pi R}(sin90-sin45)+\frac{\mu_0i}{4\pi R}(sin90+sin45)+\frac{\mu_0i}{2 R}\\=\frac{\mu_0i}{4\pi R}(\sqrt{2}+2\pi)

The direction of magnetic field due to first conductor ( in to the plane) is opposite to that of other two conductors (out of the plane).

Hence the correct option is (1)

Posted by

vishal kumar

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